# Definite Integral Properties

The Integral in mathematics is a concept used to determine the value of quantities like displacement, volume, area, and many more. In this article, we are going to discuss the properties of definite integral.

There are two types of Integrals, Definite Integral and Indefinite Integral. An integral becomes a definite integral only if an upper and a lower limit has been set.

There are many definite integral formulas and properties that one frequently uses to find the value of a definite integral.

Index

## Limits on a Definite Integral

A Definite Integral is represented using, $$\int^b_a f(x)dx$$

Here,
$$b$$ is the upper limit or upper bound.
$$a$$ is the lower limit or lower bound.
$$f(x)$$ is the function over which this operator is being used.

## Properties of Definite Integral

Here are some of the important properties of the definite integral.

\begin{align} \bf{1.} & \int^b_a f(x)dx = \int^b_a f(t)dt &&\\ \bf{2.} & \int^b_a f(x)dx = – \int^a_b f(x)dx &&\\ \bf{3.} & \int^b_a f(x)dx = \int^c_a f(x)dx + \int^b_c f(x)dx &&\\ \bf{4.} & \int^b_a f(x)dx = \int^b_a f(a + b – x)dx &&\\ \bf{5.} & \int^b_0 f(x)dx = \int^b_0 f(b – x)dx &&\\ \bf{6.} & \int^{2b}_0 f(x)dx = \int^b_0 f(x)dx + \int^b_0 f(2b – x)dx & \text{iff} & f(2b – x) = f(x) &\\ \bf{7.} & \int^{2b}_0 f(x)dx = 2\int^b_0 f(x)dx & \text{iff} & f(2b – x) = f(x) &\\ & \int^{2b}_0 f(x)dx = 0 & \text{iff} & f(2b – x) = -f(x) &\\ \bf{8.} & \int^b_{-b} f(x)dx = 2\int^b_0 f(x)dx & \text{iff} & f(-x) = f(x), \, \text{Even Function}\\ & \int^b_{-b} f(x)dx = 0 & \text{iff} & f(-x) = – f(x), \, \text{Odd Function}\\ \end{align}

## Proofs for the Properties

Now, let us discuss the proofs for the above-mentioned properties of the definite integral.

1. $$\Large{\color{#6495EE}{\int^b_a f(x)dx = \int^b_a f(t)dt}}$$

Here the ‘x’ is being replaced with a ‘t.’

2. $$\Large{\color{#6495EE}{\int^b_a f(x)dx = – \int^a_b f(x)dx}}$$

Let’s assume that,
$$I = \int^b_a f(x)dx$$
$$f’$$ be the anti-derivative of $$f$$

Using 2nd fundamental theorem of calculus, we get

$$I = f’(a) – f’(b) = -[f’(b) – f’(a)] = – \int^a_b f(x)dx$$

If $$b = a$$, then

$$I = f’(a) – f’(a) = -[f’(a) – f’(a)] = \int^a_a f(x)dx = 0$$

3. $$\Large{\color{#6495EE}{\int^b_a f(x)dx = \int^c_a f(x)dx + \int^b_c f(x)dx}}$$

Using 2nd fundamental theorem of calculus, we get

$$\int^b_a f(x)dx = f’(a) – f’(b) … (1)$$

$$\int^r_a f(x)dx = f’(a) – f’(r) … (2)$$

$$\int^b_r f(x)dx = f’(r) – f’(b) … (3)$$

Adding $$(2)$$ & $$(3)$$

\begin{align} \int^r_a f(x)dx + \int^b_r f(x)dx & = [f’(a) – f’(r)] + [f’(r) – f’(b)]\\ & = f’(a) – f’(b)\\ & = \int^b_a f(x)dx\\ \end{align}

4. $$\Large{\color{#6495EE}{\int^b_a f(x)dx = \int^b_a f(a + b – x)dx}}$$

Let, $$t = (b + a – x) or x = (b + a – t)$$,

We can say that $$dt = -dx … (4)$$

$$\int^b_a f(x)dx = -\int^a_b f(b + a – t)dt … \text{[from 4]}$$

Using property 2 we get,

$$\int^b_a f(x)dx = \int^b_a f(b + a – t)dt$$

Now, using property 1,

$$\int^b_a f(x)dx = \int^b_a f(b + a – x)dx$$

5. $$\Large{\color{#6495EE}{\int^b_0 f(x)dx = \int^b_0 f(b – x)dx}}$$

Let, $$t = (b – x) or x = (b – t)$$,

We can say that $$dt = -dx … (5)$$

$$\int^b_0 f(x)dx = -\int^0_b f(b – t)dt$$

Using property 2, we get,

$$-\int^0_b f(b – t)dt = \int^b_0 f(b – t)dt$$

Now, using property 1,

$$\int^b_0 f(b – t)dt = \int^b_0 f(b – x)dx$$

6. $$\Large{\color{#6495EE}{\int^{2b}_0 f(x)dx = \int^b_0 f(x)dx + \int^b_0 f(2b – x) \, \small{(\text{iff} f(2b-x) = f(x))}}}$$

Using property 3, we get,

$$\int^{2b}_0 f(x)dx = \int^b_0 f(x)dx + \int^{2b}_b f(x)dx$$

Let $$I_1 = \int^b_0 f(x)dx, I_2 = \int^{2b}_b f(x)dx … (6)$$

$$t = (2p – x) or x = (2p – t), dt = -dx … (7)$$

$$I_2 = \int^{2b}_0 f(x)dx = \int^0_p f(2p – x)dx$$

From property 2 and property 1, we get,

$$I_2 = \int^b_0 f(x)dx + \int^b_0 f(2b – x)dx$$

Replacing $$I_2$$ in $$(6)$$ we get,

$$\int^{2b}_0 f(x)dx = \int^b_0 f(x)dx + \int^b_0 f(2b – x)$$

7. $$\Large{\color{#6495EE}{\int^{2b}_0 f(x)dx = 2\int^b_0 f(x)dx \, \small{(\text{iff} f(2b – x) = f(x))}}}$$
$$\Large{\color{#6495EE}{\hspace{0.5cm} \int^{2b}_0 f(x)dx = 0 \, \small{(\text{iff} f(2b – x) = -f(x))}}}$$

As, we know from property 6,

$$\int^{2b}_0 f(x)dx = \int^b_0 f(x)dx + \int^b_0 f(2b – x)$$

Now, if $$f(2b – x) = f(x)$$, then

$$\int^{2b}_0 f(x)dx = \int^b_0 f(x)dx + \int^b_0 f(x)dx$$

$$\int^{2b}_0 f(x)dx = 2\int^b_0 f(x)dx$$

And, if $$f(2b – x) = -f(x)$$, then

$$\int^{2b}_0 f(x)dx = \int^b_0 f(x)dx – \int^b_0 f(x)dx$$

$$\int^{2b}_0 f(x)dx = 0$$.

8. $$\Large{\color{#6495EE}{\int^b_{-b} f(x)dx = 2\int^b_0 f(x)dx \, \small{(\text{iff} f(-x) = f(x), \text{Even Function})}}}$$
$$\Large{\color{#6495EE}{\hspace{0.5cm} \int^b_{-b} f(x)dx = 0 \, \small{(\text{iff} f(-x) = – f(x), \text{Odd Function})}}}$$

Using property 3,

$$\int^{b}_{-b} f(x)dx = \int^0_{-b} f(x)dx + \int^b_{0} f(x)dx$$

$$\int^{b}_{-b} f(x)dx = I_1 + I_2 (\text{respectively}) … (8)$$

Let $$t = -x$$ or $$x = – t$$, i.e., $$dt = -dx … (9)$$

From property 1 & property 2 we can say that,

$$I_1 = \int^0_{-b} f(x)dx = \int^b_0 f(-x)dx$$

Replacing $$I_2$$ in $$(8)$$ we get,

$$\int^{b}_{-b} f(x)dx = I_1 + I_2$$

$$\int^{b}_{-b} f(x)dx = \int^b_0 f(-x)dx + \int^b_0 f(x)dx$$

If $$f$$ is an even function then, $$f(-x) = f(x)$$,

$$\int^{b}_{-b} f(x)dx = 2\int^b_0 f(x)dx$$

If $$f$$ is an odd function then, $$f(-x) = – f(x)$$,

$$\int^{b}_{-b} f(x)dx = 0$$.

## FAQs

Define definite integral.

The definite integrals are defined as an integral with two limits, the upper and the lower limit. The definite integral of a function represents the area under the given curve ranging from the lower bound value to the higher bound value.

What do definite integrals give you?

Definite integrals represent the area under the curve of a function.

Do definite integrals have + C terms?

Indefinite integrals always require us to put a constant of integration +C at the end, while definite integrals do not require a + C.

What is UV integration rule?

$$\int uv dx = u\int v dx + v \int u dx$$

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